Example №1. Solving of a System of Linear Equations by the Gauss Elimination (One Solution)This solution was made using the calculator presented on the site. Example №2. Solving of a system of linear equations by the Gauss elimination (many solutions) Example №3. Solving of a system of linear equations by the Gauss elimination (no solution) Example №4. Solving of a system of linear equations by the Gauss Jordan elimination (one solution) Example №5. Solving of a system of linear equations by the Gauss Jordan elimination (many solutions) Please note that the coefficients will disappear which located in the "red" positions.
The equation 1 multiplied by 1 is added to the equation 2. more info ( 5 x_{1} + 4 x_{1} * ( 1) ) + ( 3 x_{2} + 2 x_{2} * ( 1) ) + ( 5 x_{3} + ( 3 x_{3}) * ( 1) ) = 8 + ( 3) * ( 1) This transformation will allow us to count without fractions for some time.
The equation 2 and equation 1 are reversed.
The equation 1 multiplied by 4 is added to the equation 2. more info ( 4 x_{1} + x_{1} * ( 4) ) + ( 2 x_{2} + x_{2} * ( 4) ) + ( 3 x_{3} + ( 2 x_{3}) * ( 4) ) = 3 + ( 5) * ( 4) The "red" coefficient is zero.
The equation 1 multiplied by 4 is added to the equation 3. more info ( 4 x_{1} + x_{1} * ( 4) ) + ( x_{2} + x_{2} * ( 4) ) + ( 5 x_{3} + ( 2 x_{3}) * ( 4) ) = 22 + ( 5) * ( 4) The "red" coefficient is zero.
The equation 3 multiplied by 1 is added to the equation 2. more info ( 2 x_{2} + ( 3 x_{2}) * ( 1) ) + ( 5 x_{3} + 13 x_{3} * ( 1) ) = 17 + 42 * ( 1) This transformation will allow us to count without fractions for some time.
The equation 2 multiplied by 3 is added to the equation 3. more info ( 3 x_{2} + x_{2} * 3 ) + ( 13 x_{3} + ( 8 x_{3}) * 3 ) = 42 + ( 25) * 3 The "red" coefficient is zero.
We will find the variable x_{3} from equation 3 of the system.  11 x_{3} =  33 x_{3} = 3 We will find the variable x_{2} from equation 2 of the system. x_{2}  8 x_{3} =  25 x_{2} =  25 + 8 x_{3} x_{2} =  25 + 8 * ( 3 ) x_{2} =  1 We will find the variable x_{1} from equation 1 of the system. x_{1} + x_{2}  2 x_{3} =  5 x_{1} =  5  x_{2} + 2 x_{3} x_{1} =  5  (  1 ) + 2 * ( 3 ) x_{1} = 2 Result: x_{1} = 2 x_{2} =  1 x_{3} = 3
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